Lab 02: Uniformly Accelerated Motion

Learning Objectives

  • Calculate the acceleration of a body using a distance-versus-time-squared graph.
  • Interpret the motion of an object using distance-versus-time, distance-versus-time-squared, velocity-versus-time, and acceleration-versus-time graphs.
  • Determine average velocity, instantaneous velocity, average acceleration, and instantaneous acceleration from measured motion data.
  • Compare graphical and analytical methods for describing uniformly accelerated motion.
  • Identify common experimental errors in timing, distance measurement, release method, and rail alignment.

This experiment studies the motion of a trolley moving along an inclined rail. When the component of gravity along the rail is nearly constant and friction is small, the trolley accelerates at an approximately constant rate. By measuring the distance traveled and the time of travel, students can analyze motion using tables, formulas, and graphs.

Target Learning Outcome

TLO 2: Calculate the acceleration of a body based on a distance-versus-time-squared graph of its motion, and interpret the motion of an object given a distance-versus-time or distance-versus-time-squared graph.

I. Discussion of Theory

Uniformly accelerated motion

Uniformly accelerated motion occurs when an object changes velocity by equal amounts during equal time intervals. In this experiment, the trolley starts from rest and moves down a slightly inclined rail. The acceleration is not expected to be exactly equal to gg because only the component of gravity along the incline accelerates the trolley, and friction opposes the motion.

Motion

Motion is the change in position of an object with respect to a chosen reference point over time.

Distance

Distance is a scalar quantity representing the total length of the path traveled by an object, regardless of its direction.

Displacement

Displacement is a vector quantity representing the straight-line distance and direction from an object's initial position to its final position.

Average Velocity

Average velocity is the displacement divided by the elapsed time interval.

Average velocity

Average velocity over a time interval is computed from the change in displacement divided by the change in time.

vˉ=ΔdΔt=d2d1t2t1\bar{v} = \frac{\Delta d}{\Delta t} = \frac{d_2 - d_1}{t_2 - t_1}

Variables

SymbolDescriptionUnit
vˉ\bar{v}average velocitym/s or cm/s
Δd\Delta dchange in displacement or distance traveled along the railm or cm
Δt\Delta telapsed time intervals

Instantaneous Velocity

Instantaneous velocity is the velocity of the object at a particular instant. On a distance-versus-time graph, it is represented by the slope of the tangent line at that instant.

Average Acceleration

Average acceleration is the change in velocity divided by the elapsed time interval.

Average acceleration

Average acceleration is computed from the change in velocity divided by the change in time.

aˉ=ΔvΔt=v2v1t2t1\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}

Variables

SymbolDescriptionUnit
aˉ\bar{a}average accelerationm/s2orcm/s2m/s^2 or cm/s^2
Δv\Delta vchange in velocitym/s or cm/s
Δt\Delta telapsed time intervals

Instantaneous Acceleration

Instantaneous acceleration is the acceleration at a particular instant. On a velocity-versus-time graph, it is represented by the slope of the tangent line at that instant.

Constant acceleration kinematics

For motion with constant acceleration, displacement depends on initial velocity, time, and acceleration.

d=v0t+12at2d = v_0t + \frac{1}{2}at^2

If the trolley starts from rest, then v0=0v_0 = 0, so:

d=12at2d = \frac{1}{2}at^2

Variables

SymbolDescriptionUnit
dddistance traveled along the railm or cm
v0v_0initial velocitym/s or cm/s
aaaccelerationm/s2orcm/s2m/s^2 or cm/s^2
tttimes

Why graph distance versus time squared?

For a trolley released from rest under approximately constant acceleration,

d=12at2d = \frac{1}{2}at^2

This has the same form as a straight-line equation, y=mxy = mx, where y=dy = d, x=t2x = t^2, and the slope is a/2a/2. Therefore:

a=2(slope of the d vs. t2 graph)a = 2(\text{slope of the } d \text{ vs. } t^2 \text{ graph})

A straight-line dd versus t2t^2 graph is strong evidence that the motion is uniformly accelerated.

Expected Trends

When an object undergoes uniformly accelerated motion starting from rest:

  • Distance versus time (dd vs tt) graph: The plotted points should form a curve that opens upward (a parabola), indicating that the object covers increasing distances in equal time intervals as it speeds up.
  • Distance versus time squared (dd vs t2t^2) graph: The plotted points should form an approximately straight line passing through the origin. This linearity strongly supports the relationship d=12at2d = \frac{1}{2}at^2.
  • Velocity versus time (vv vs tt) graph: The plotted points should form an approximately straight line with a constant positive slope, indicating a constant rate of acceleration.

II. Equipment / Materials Needed

Equipment or materialPurpose
TrolleyMoving object whose motion will be measured.
Metal railTrack on which the trolley moves.
Iron standSupports one end of the rail to create an incline.
StopwatchMeasures travel time.
MeterstickMeasures distance traveled along the rail.
Ruler or release stickTemporarily holds the trolley before release.
Masking tape or markersMarks selected distance positions along the rail.
Graphing paper or spreadsheetUsed for plotting motion graphs.

Alternative setups

The trolley and metal rail may be replaced by a linear air track and rider, a wooden plane and a freely running toy car, or a simplified Atwood's machine. The important requirement is that the object must move with approximately constant acceleration.

Safety and setup reminders

Keep the rail stable on the iron stand. Make sure the trolley cannot fall off the table or hit another group. Do not raise the rail too steeply because the trolley may move too fast for accurate stopwatch timing.

III. Diagram of Setup

Suggested apparatus arrangement

            Iron stand
               |
               |        raised end of rail
               |          _________
               |         /        /|
               |        / Trolley/ |
               |_______/________/  |
                      /        /   |
                     / Metal /    |
                    / Rail  /     |
                   /______/______|
                 lower end / finish mark

Measure distance d along the rail from the release point to the selected mark.

IV. Procedures

Experimental procedure

  1. Set up the rail on the iron stand so that the rail forms a small incline. The lower end of the rail should be stable and clear of obstacles.
  2. Mark several distances along the rail from the release point. Suggested positions are d0d_0, d1d_1, d2d_2, d3d_3, and d4d_4.
  3. Place the trolley at the release point and hold it using a ruler, stick, or gate. Avoid pushing the trolley.
  4. Remove the ruler or stick to release the trolley and start the stopwatch at the same instant.
  5. Stop the stopwatch when the trolley reaches the selected lower mark or desired displacement.
  6. Record the distance traveled dd and time tt in Table 2.1.
  7. Repeat the timing for each distance. If time permits, take at least three trials per distance and use the average time.
  8. Compute the average velocity for each time interval and record the values in Table 2.2.
  9. Plot distance traveled dd as the ordinate and time tt as the abscissa. Draw a smooth curve through the mean of the points.
  10. From the distance-versus-time graph, estimate the instantaneous velocity of the trolley at each selected time using tangent slopes. Record the values in Table 2.3.
  11. Compute the average acceleration for each time interval and record the values in Table 2.4.
  12. Plot instantaneous velocity as the ordinate and time tt as the abscissa. Draw a smooth best-fit line or curve through the points.
  13. From the velocity-versus-time graph, estimate the instantaneous acceleration at each selected time using tangent slopes. Record the values in Table 2.5.
  14. Plot distance dd versus t2t^2. Determine the slope and compute the acceleration using a=2(slope)a = 2(\text{slope}).

Release method

The trolley must be released, not pushed. A push gives the trolley an unknown initial velocity, which changes the expected relationship from d=12at2d = \frac{1}{2}at^2 to d=v0t+12at2d = v_0t + \frac{1}{2}at^2.

V. Student Information

FieldEntry
Name
Schedule
Group No.
Date Performed

VI. Data and Results

Table 2.1. Distance Traveled dd versus Time tt

Data pointDistance traveled, dd (cm)Time, tt (s)Time squared, t2t^2 (s²)
0d0=d_0 =t0=t_0 =t02=t_0^2 =
1d1=d_1 =t1=t_1 =t12=t_1^2 =
2d2=d_2 =t2=t_2 =t22=t_2^2 =
3d3=d_3 =t3=t_3 =t32=t_3^2 =
4d4=d_4 =t4=t_4 =t42=t_4^2 =

Table 2.2. Average Velocity during Each Time Interval

Time intervalFormulaAverage velocity, cm/s
t0t_0 to t1t_1d1d0t1t0\frac{d_1-d_0}{t_1-t_0}
t1t_1 to t2t_2d2d1t2t1\frac{d_2-d_1}{t_2-t_1}
t2t_2 to t3t_3d3d2t3t2\frac{d_3-d_2}{t_3-t_2}
t3t_3 to t4t_4d4d3t4t3\frac{d_4-d_3}{t_4-t_3}

Table 2.3. Instantaneous Velocity at Selected Times

TimeMethodInstantaneous velocity, cm/s
t0t_0Slope of tangent to dd-tt graph
t1t_1Slope of tangent to dd-tt graph
t2t_2Slope of tangent to dd-tt graph
t3t_3Slope of tangent to dd-tt graph
t4t_4Slope of tangent to dd-tt graph

Table 2.4. Average Acceleration during Each Time Interval

Time intervalFormulaAverage acceleration, cm/s²
t0t_0 to t1t_1v1v0t1t0\frac{v_1-v_0}{t_1-t_0}
t1t_1 to t2t_2v2v1t2t1\frac{v_2-v_1}{t_2-t_1}
t2t_2 to t3t_3v3v2t3t2\frac{v_3-v_2}{t_3-t_2}
t3t_3 to t4t_4v4v3t4t3\frac{v_4-v_3}{t_4-t_3}

Table 2.5. Instantaneous Acceleration at Selected Times

TimeMethodInstantaneous acceleration, cm/s²
t0t_0Slope of tangent to vv-tt graph
t1t_1Slope of tangent to vv-tt graph
t2t_2Slope of tangent to vv-tt graph
t3t_3Slope of tangent to vv-tt graph
t4t_4Slope of tangent to vv-tt graph

Table 2.6. Distance versus Time Squared Graph Data

Data pointdd (cm)t2t^2 (s²)Graphing note
0Plot as (t02,d0)(t_0^2, d_0)
1Plot as (t12,d1)(t_1^2, d_1)
2Plot as (t22,d2)(t_2^2, d_2)
3Plot as (t32,d3)(t_3^2, d_3)
4Plot as (t42,d4)(t_4^2, d_4)

Table 2.7. Graph Slope and Acceleration Summary

GraphMeasured SlopeComputed AccelerationExpected Trend Match?
dd vs t2t^2
vv vs tt

Graphing Rubric

Graphing requirements

VII. Computations

Required computations

  1. Compute t2t^2 for every measured time value.
  2. Compute average velocity for each interval using vˉ=Δd/Δt\bar{v}=\Delta d/\Delta t.
  3. Determine instantaneous velocity from the slope of the tangent line on the dd versus tt graph.
  4. Compute average acceleration for each interval using aˉ=Δv/Δt\bar{a}=\Delta v/\Delta t.
  5. Determine instantaneous acceleration from the slope of the tangent line on the vv versus tt graph.
  6. Plot dd versus t2t^2 and find the slope of the best-fit line.
  7. Compute the experimental acceleration using a=2(slope)a = 2(\text{slope}).
  8. State whether the dd versus t2t^2 graph supports uniformly accelerated motion.

Average Velocity

Computed as the change in distance over the change in time.

vˉ=ΔdΔt\bar{v} = \frac{\Delta d}{\Delta t}

Instantaneous Velocity

Calculated as the slope of the tangent line on a distance-versus-time graph.

v=slope of d vs. tv = \text{slope of } d \text{ vs. } t

Average Acceleration

Computed as the change in velocity over the change in time.

aˉ=ΔvΔt\bar{a} = \frac{\Delta v}{\Delta t}

Instantaneous Acceleration

Calculated as the slope of the tangent line on a velocity-versus-time graph.

a=slope of v vs. ta = \text{slope of } v \text{ vs. } t

Displacement for Uniform Acceleration

General formula and formula for release from rest (v0=0v_0 = 0).

d=v0t+12at2d = v_0t + \frac{1}{2}at^2d=12at2d = \frac{1}{2}at^2

Acceleration from the dd versus t2t^2 graph

The graph slope gives one-half of the acceleration when the object starts from rest.

slope=ΔdΔ(t2)=12a\text{slope} = \frac{\Delta d}{\Delta(t^2)} = \frac{1}{2}aa=2ΔdΔ(t2)a = 2\frac{\Delta d}{\Delta(t^2)}

Variables

SymbolDescriptionUnit
aaexperimental accelerationcm/s2orm/s2cm/s^2 or m/s^2
Δd\Delta dchange in distance between two points on the best-fit linecm or m
Δ(t2)\Delta(t^2)change in time squared between the same two pointss2s^2

Computing time squared

If the measured travel time is t=1.40st = 1.40\,\text{s}, the time squared is computed as:

t2=(1.40s)2=1.96s2t^2 = (1.40\,\text{s})^2 = 1.96\,\text{s}^2

Computing average velocity

Suppose the trolley travels from d1=10cmd_1 = 10\,\text{cm} at t1=1.0st_1 = 1.0\,\text{s} to d2=40cmd_2 = 40\,\text{cm} at t2=2.0st_2 = 2.0\,\text{s}. The average velocity during this interval is:

vˉ=40102.01.0=301.0=30cm/s\bar{v} = \frac{40 - 10}{2.0 - 1.0} = \frac{30}{1.0} = 30\,\text{cm/s}

Computing acceleration from d vs t² graph

Suppose two points from the best-fit dd versus t2t^2 graph are (0.80s2,20cm)(0.80\,\text{s}^2, 20\,\text{cm}) and (2.40s2,60cm)(2.40\,\text{s}^2, 60\,\text{cm}).

slope=60202.400.80=401.60=25cm/s2\text{slope} = \frac{60 - 20}{2.40 - 0.80} = \frac{40}{1.60} = 25\,\text{cm/s}^2a=2(25)=50cm/s2a = 2(25) = 50\,\text{cm/s}^2

Therefore, the experimental acceleration is 50cm/s250\,\text{cm/s}^2 or 0.50m/s20.50\,\text{m/s}^2.

VIII. Observations and Conclusions

Observation guide

Error Analysis and Accuracy Improvement

Experimental errors can significantly affect the calculated acceleration. Common sources of error include:

  • Timing delays: Human reaction time when starting or stopping the stopwatch can introduce significant errors, especially for short distances. This can be mitigated by taking multiple trials and averaging the times.
  • Release method: Pushing the trolley instead of releasing it from rest introduces an unknown initial velocity, invalidating the assumption that v0=0v_0 = 0.
  • Rail alignment: An uneven or warped rail causes non-uniform acceleration. Ensure the rail is straight and stable.
  • Distance measurement: Inaccurate markings or parallax error when reading the meterstick can affect distance values.

Conclusion guide

A good conclusion should state the measured acceleration of the trolley, explain how the graphs support or do not support uniformly accelerated motion, and identify the largest likely source of experimental error.

Lab Report Format

Your final lab report should include the following sections:

  1. Title Page: Experiment title, your name, group members, schedule, and date performed.
  2. Objectives: Briefly restate the goals of the experiment in your own words.
  3. Data and Results: Include all completed tables (Tables 2.1 to 2.7) and the three required graphs (dd vs tt, vv vs tt, dd vs t2t^2).
  4. Computations: Show complete sample calculations for one set of data points (e.g., computing t2t^2, average velocity, and experimental acceleration).
  5. Observations and Error Analysis: Discuss the trends seen in your graphs and analyze possible sources of experimental error.
  6. Conclusion: State the final calculated acceleration, verify if the motion was uniformly accelerated, and directly address the target learning outcome.

IX. Questions and Problems

  1. Define the following terms:

    1. Motion
    2. Average velocity
    3. Instantaneous velocity
    4. Average acceleration
    5. Instantaneous acceleration
  2. The position of a particle moving along the xx-axis is given in meters by

x=9.75m+(1.5ms3)t3x = 9.75\,\text{m} + \left(1.5\,\frac{\text{m}}{\text{s}^3}\right)t^3

where tt is in seconds. Calculate:

  1. the average velocity during the time interval t=3.00st = 3.00\,\text{s} to t=5.00st = 5.00\,\text{s};

  2. the instantaneous velocity at t=3.00st = 3.00\,\text{s};

  3. the instantaneous velocity at t=6.00st = 6.00\,\text{s};

  4. the average acceleration from t=3.00st = 3.00\,\text{s} to t=6.00st = 6.00\,\text{s};

  5. the instantaneous acceleration at t=2.00st = 2.00\,\text{s}.

  6. A record of travel along a straight path is as follows:

    1. Start from rest with a constant acceleration of 2.77m/s22.77\,\text{m/s}^2 for 15.0s15.0\,\text{s}.
    2. Maintain a constant velocity for the next 2.05min2.05\,\text{min}.
    3. Apply a constant negative acceleration of 9.47m/s29.47\,\text{m/s}^2 for 4.39s4.39\,\text{s}.
    4. What was the total displacement for the trip?
    5. What were the average speeds for parts a, b, and c of the trip, as well as for the complete trip?

Selected Answer Key

Problem 2:

  1. Average velocity: 73.5m/s73.5\,\text{m/s}
  2. Instantaneous velocity at t=3t=3: 40.5m/s40.5\,\text{m/s}
  3. Instantaneous velocity at t=6t=6: 162m/s162\,\text{m/s}
  4. Average acceleration: 40.5m/s240.5\,\text{m/s}^2
  5. Instantaneous acceleration at t=2t=2: 18m/s218\,\text{m/s}^2

Problem 3: 4. Total displacement: 5513.44m5513.44\,\text{m} (or 5.51km5.51\,\text{km}) 5. Average speeds: 20.78m/s20.78\,\text{m/s}, 41.55m/s41.55\,\text{m/s}, 20.77m/s20.77\,\text{m/s}, and 38.72m/s38.72\,\text{m/s}

X. References

Bueche, F. J., & Hecht, E. (1997). Schaum's Outline of Theory and Problems of College Physics (9th ed.). New York: McGraw-Hill.

Instructor note

The original HTML worksheet has been converted into MDX and expanded with the missing theory, graph interpretation guide, computation formulas, data table improvements, and conclusion prompts. The interactive HTML theme toggle and input fields were intentionally converted into printable MDX tables so this page works consistently with the existing content renderer.